How do you find a unit vector normal to the surface #x^3+y^3+3xyz=3# ay the point(1,2,-1)?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

23
Oct 25, 2016

Answer:

#{-1/sqrt[14], 3/sqrt[14], sqrt[2/7]}#

Explanation:

Calling

#f(x,y,z)=x^3+y^3+3xyz-3=0#

The gradient of #f(x,y,z)# at point #x,y,z# is a vector normal to the surface at this point.

The gradient is obtained as follows

#grad f(x,y,z) = (f_x,f_y,f_z) = 3(x^2+yz,y^2+xz,xy)# at point
#(1,2,-1)# has the value
#3(-1,3,2)# and the unit vector is
#({-1,3,2})/sqrt(1+3^2+2^2)={-1/sqrt[14], 3/sqrt[14], sqrt[2/7]}#

Was this helpful? Let the contributor know!
1500